Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible coordinate of vertex R? A. (-3, -6)
B. (0, 8)
C. (1, 8.9)
D. (1, -8.9)

Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

By simplifying it we get,

Now,

again finding the distance between PR,

or,

By simplifying it we get,

now, finding the distance of QR,

or, by simplification we get,

now, equating PR and QR,

we cancelled the root ,

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

cancel the roots,

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

Hope it helps...

80° is the measure of gh

step-by-step explanation:

circle f is congruent to circle j

we have two circle with center f and j.

both circles are congruent. the radius of the circle would be equal.

see the attachment for the circle.

we are given ∠efd ≅ ∠gjh

therefore, arc (ed) = arc (gh)     (∴ circle f is congruent to circle j)

but ∠dfe=80°=arc (ed)

therefore, arc (gh) = 80°

thus, 80° is the measure of gh

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct